1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding the focus of a parabola given an equation

  1. Mar 26, 2016 #1
    1. The problem statement, all variables and given/known data
    Alright, so the equation of a parabola is [itex]y = 1/4p*x^2[/itex], P being either an x or y value, and the other x or y being zero. Let's say that [itex]x^2 = 16y[/itex]. If you divide both sides by 16, you get [itex]y = x^2/16[/itex], which can be simplified to [itex]y = 1/16*x^2[/itex]. This is in the format of a parabola, so finding p is simple. 16 is the product of 4p, so [itex]4p=16[/itex]. Divide both sides by 4 and p=4. So the focus is at (0,4).

    This seems simple to me, until you get equations like this: [itex]3x^2 + 4y = 0[/itex]. I can't seem to get this in the form of a parabola. I got [itex]-3/4 *x^2=y[/itex]. This wouldn't help me find the focus, due to the fact that it's not in that form. How exactly would you go about finding the focus in cases like this? Thanks for any help!
    Last edited by a moderator: Mar 26, 2016
  2. jcsd
  3. Mar 26, 2016 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Hello BlueQuark. Welcome to PF !

    You need parentheses around the 4p since both are in the denominator. You are using LaTeX, so it is easy to use "\frac" to write : ##\ y = \frac{1}{4p}x^2\ ##.

    Writing the equation of the parabola as ##\ 4py = x^2\ ## may prove to be even handier for solving your problem.
  4. Mar 26, 2016 #3


    User Avatar
    Homework Helper

    4y + 3x^2 = 0
    <=> y = -3/4 x^2
    <=> y = -1/(4/3)x^2

    I don't know whether this has answered your question.
  5. Mar 26, 2016 #4


    User Avatar
    Science Advisor

    As SammyS said, that should be [itex]y= 1/(4p)x^2[/itex] or [itex]y= x^2/(4p)[/itex]

    I really don't know what that means. What does it mean to say that p is "either an x or y value". What do you mean by an "x value" or a "y value"? And what "other" x or y do you mean? p is, of course, the x coordinate of the focus- in this situation, the focus is at (0, p). Is that what you mean?

    Surely you know that [itex]-3/4= \frac{1}{-\frac{4}{3}}[/itex]? [itex]p= -\frac{4}{3}[/itex] and the focus is at [itex]\left(0, -\frac{4}{3}\right)[/itex].
  6. Mar 27, 2016 #5


    User Avatar
    Homework Helper

    Yes, [itex]-3/4= \frac{1}{-\frac{4}{3}}[/itex]

    And, y = -x^2/(4p) is a valid equation for a parabola.
    P = -1/3, not -4/3 so the focus is at (0, -1/3)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted